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Thread: Use Multi-Section Windings to Reduce Self-Capacitance

  1. #36
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    Quote Originally Posted by Bagpipe View Post
    mike: Flux created by Current. I have to let that sag. Sorry... i don't seem to understand your words, nor the sense of your words. You have a static magnetic field - and you have a coil (or a split-coil) inside it. Do we still talk about the same issue here? Or are you ran away into transformers?
    The static magnetic field is irrelevant in this discussion. The varying magnetic field created by ac current in the coil is what is responsible for inductance. the more this flux (field time area) is shared among turns the higher the inductance,

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    Quote Originally Posted by Bagpipe View Post
    Antigua:
    What happens to the resonance peak with increased series resistance OR increased Eddy Currents (almost the same)??
    Where do eddy current enter into the picture? The effects are similar but not identical.

    Quote Originally Posted by Bagpipe View Post
    Winding to Winding magnetic coupling? What are you talking about? There is no drop in inductance. Simple as that. A Split Bobbin Inductor has almost EXACTLY the same inductance as a single bobbin inductor. WHY should inductance fall?
    The increase in inductance with added turns of wire does not increase at a linear rate, it's square the number of turns, due to magnetic coupling between the turns. If you pull the coil into two or more parts, you break up that coupling. You end up with two inductors, both having somewhat less than half the inductance of the original whole.


    Quote Originally Posted by Bagpipe View Post
    When you turn the volume down on your guitar, what happens to series resistance seen by the amp? And what happens to miller effect?
    I don't know, you tell me.

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  3. #38
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    Quote Originally Posted by Joe Gwinn View Post
    One can compensate for reduced winding space by using the next finer wire size. Unless the separators are fairly thick, inductance won't be much affected, so long as the same total number of turns is used. Cable capacitance is of course unaffected, but reducing the self-capacitance of the pickup increases our wiggle room.
    I do not understand how you are so confident that the inductance does not change. Have you done this, and measured the result with an air core, or short cores as in a pickup?

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    Last edited by Bagpipe; 05-21-2018 at 04:34 AM.

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    Last edited by Bagpipe; 05-21-2018 at 04:34 AM.

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    Quote Originally Posted by Bagpipe View Post
    Please read again, what you write. It is nonsense. Really. Not because i dislike you, but beacuse it is utter nonsense.

    The varying mag. field is in truth a static magnetic field, which is disturbed by the string. This disturbance causes a disturbance in the coil - more turns - more pronounciated.

    Current plays absolutely NO ROLE - except we talk of the effects of the load.

    Sorry Mike - you must know, that i bear GREAT RESPECT for you. You may not be aware of it, but your publications are with me for at least 3 years now. I regard you as a star in the scene. But with this topic here, i just lose understanding for your points. Sorry. I don't want to offend you.
    From inductance, Wikipedia:

    Inductance is a property of an electrical conductor which opposes a change in current.[1] It does that by storing and releasing energy from a magnetic fieldsurrounding the conductor when current flows, according to Faraday's law of induction. When current rises, energy (as magnetic flux) is stored in the field, reducing the current and causing a drop in potential (i.e., a voltage) across the conductor; when current falls, energy is released from the field supplying current and causing a rise in potential across the conductor. Mutual inductance describes the change of current in a circuit when a second circuit also experiences a change of current; energy is coupled from one circuit to the other through magnetic fields.
    These effects are derived from two fundamental observations of physics: a steady current creates a steady magnetic field described by Oersted's law,[2] and a time-varying magnetic field induces an electromotive force (EMF) in nearby conductors, which is described by Faraday's law of induction.[3] According to Lenz's law,[4] a changing electric current through a circuit that contains inductance induces a proportional voltage, which opposes the change in current (self-inductance). The varying field in this circuit may also induce an EMF in neighbouring circuits (mutual inductance).
    The circuit component representing inductance is called an inductor. The term inductance was coined by Oliver Heaviside in 1886.[5]

    Pickups cannot be understood by this "disturb the static magnetic field" thing. You need to apply Faraday's law. For example, check MacDonald (Princeton).

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  7. #42
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    Quote Originally Posted by Bagpipe View Post
    Please read again, what you write. It is nonsense. Really. Not because i dislike you, but beacuse it is utter nonsense.

    The varying mag. field is in truth a static magnetic field, which is disturbed by the string. This disturbance causes a disturbance in the coil - more turns - more pronounciated.
    The residual flux of the AlNiCo or ceramic does not change as the string moves around, it is not "disturbed". Voltage is caused by magnetic change through the winds of the coil, and the primary agent of magnetic change is the moving, magnetized guitar string. The permeability of the screws, slugs or pole pieces will increase the total amount of change, but that change starts with the magnetized guitar string.

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    Last edited by Bagpipe; 05-21-2018 at 04:35 AM.

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    Quote Originally Posted by Bagpipe View Post
    They are identical - except for the frequency dependence of eddies.

    This is Transformer-Speak. Lets get back to the basics. We have an inductive sensor here. It is not just: Look the other way around at it. It is not that easy.

    No - i don't. Sorry - i am no answering questions machine. And double *NOT* with people, who play - but not *are* true with their questions.
    These read like non sequiturs.

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    Last edited by Bagpipe; 05-21-2018 at 04:35 AM.

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    Quote Originally Posted by Antigua View Post
    These read like non sequiturs.
    There is a simple cure: Close your Browser and do not read it, if you don't like it.

    Miller Effect is the A - of the EE ABC. If you are not aware of it, i cannot help you.

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    Quote Originally Posted by Bagpipe View Post
    Mike: You cannot apply wiki-description here. You don't send current through the coil and observe inductance or the field. It is the other way around. This is no Transformer. And Still: I don't get your picture-.
    The magnetic field from the vibrating magnetized string induces a voltage in series with the coil. Current flows because of the load, including the pickup self capacitance as part of the load. When current flows, the inductance of the coil tends to reduce temporal changes in the current just like any other inductor.

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    Quote Originally Posted by Bagpipe View Post
    Residual Flux?

    You mean total flux? Or what? I don't get it.

    As i see it:

    A static magnetic field is put into motion (changing magnetic field) by the string. That is exactly what you said, what i said. Where is current in this picture (assuming lossless load)
    In order for the static magnetic field to be put into motion, the magnet itself would have to start moving around. Since that obviously doesn't happen, it's easy to rule out your mental model.

    If you're a coil, and you're waiting around for magnetic change to happen, you will see most of that change come from the guitar string, as it moves around. The amount of magnetic in the pole pieces will be a function of the permeability of the metal. So for example, if the core was a ceramic magnet, it might have a high Br, but it has almost no permeability at all, so you, the coil, would exclusively see magnetic change from the guitar string, and non from the ceramic core.

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    Quote Originally Posted by Bagpipe View Post
    Miller Effect is the A - of the EE ABC. If you are not aware of it, i cannot help you.
    This is just a red herring. You're saying intrinsic capacitance with the volume turned down is irrelevant because of Miller effect, but you're not going into any detail at all as to why.

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    Last edited by Bagpipe; 05-21-2018 at 04:35 AM.

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    Quote Originally Posted by Antigua View Post
    This is just a red herring. You're saying intrinsic capacitance with the volume turned down is irrelevant because of Miller effect, but you're not going into any detail at all as to why.
    I went into detail. Read my words.

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    Quote Originally Posted by Bagpipe View Post
    I went into detail. Read my words.
    I double checked; no detail.

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    Last edited by Bagpipe; 05-21-2018 at 04:35 AM.

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    Last edited by Bagpipe; 05-21-2018 at 04:35 AM.

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    Quote Originally Posted by Bagpipe View Post
    Lets say, i can follow you. Lets say, i do not speak of magnetized strings, but of strings made of permeable material. Yes: Assuming a load causes current to flow - sure. And when current flows, inductor action takes place. Clear.

    Now, where exactly does a split bobbin reduce inductance? Given, that all other parameters such as magnet and turns are equal?

    Sorry: I don't want to bother you, even if it seems that way - i REALLY do not want to bother you. I just have an understanding problem?

    Turn the question around: "Have YOU ever experimented with split bobbins, equal number of turns and according measurements?"

    I have to admit, that my knowledge about this is theoretical.
    I did not say it reduces inductance; I said I do not know if it changes it significantly, or which way if it does. The reason for a change, if any, is that in going to a split coil you are at least to some extent changing how much flux generated by current in any turn passes through all the other turns. It is clear that you do not understand how an inductor works, and so you need to study some physics and electronics.

    If a string is made of permeable material, then it becomes magnetized when placed in a static magnetic field. (While it is in the field)

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    Last edited by Mike Sulzer; 05-20-2018 at 09:09 PM.

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    Last edited by Bagpipe; 05-21-2018 at 04:36 AM.

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    Quote Originally Posted by Bagpipe View Post
    Then check again.

    I talked about your Volume Pot. This as a small hint.
    I'm calling it; there's nothing to this.

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    Last edited by Bagpipe; 05-21-2018 at 04:36 AM.

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    Quote Originally Posted by Bagpipe View Post
    With "moving" i meant: "Flux in Motion" not spinning magnets, my friend. And: Regarding your thesis about Ferrites having almost no permeability: I simply forget it as fast as i read it. Sorry... This is too far out for me. I cannot answer this.
    Who said anything about "spinning" magnets? I'll admit, I don't know all there is to know about electrical theory, but I think it's time you admit the same, you're presuming you understand the issues at hand more than you really do.

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    Last edited by Bagpipe; 05-21-2018 at 04:36 AM.

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    Quote Originally Posted by Bagpipe View Post
    How much Flux is generated by current? And: I don't know, if that is your "retoure-hit" at me, but: Telling somebody, he is stupid, wouldn't improve things at all. I already apologized to you for my language. I do it again. So: Can we now move over to EE again? I did't want to hurt you, if it is that, what you want to tell me.

    Please help me again, in understanding your thoughts.

    We are so far accord with the fact, that the magnetized/or permeable guitar string does something to the field. My understanding is, that the string changes the properties of an otherwise static magnetic field. I always thought, that this change is being picked up by the coil. You tell me, i am wrong. I am worried. Please explain in more detail, where i am wrong and why.

    Thank you very much.

    And: I would be very thankful, if you would consider giving me another chance of a friendly conversation between us two.

    Thanks again.
    The field from the magnetized string adds to the original field from the permanent magnet. But it is better to forget about the original static field from the magnet and just concentrate on the part from the magnetized string, because it is what varies with the string vibration, and thus causes changing flux through the pickup coil, which then induces the voltage by Faraday's law. We can do this because fields can be considered the sums of other fields as long as no material is close to saturation, as could happen in an audio output transformer, for example.

    This is the last reply I can make today because I have to leave this computer, and my laptop is broken.

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  27. #62
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    Quote Originally Posted by Bagpipe View Post
    And also to you Antigua: Sorry for the escalation. I am sure, you are a pretty cool guy. Was my fault. Sorry. I should have behaved better. You are right.

    Back to your thesis: Ever considered AlNiCo Magnet Poles in your explanation above?

    I understand, that you point out, that a permanent magnet hasn't really a "permeability". It is a permanent magnet.

    I also understand that a Ferrite (non-permanent magnet) has huge permeability, but no own magnetism in a sense of a permanent magnet.

    But: Does this play a role here? The poles just conduct the flux of the permanent at the bottom. You could also omit the ceramic on the bottom and use -whatever- polepieces you like. AlNico or Ceramic - as far as i know, this plays only a secondary role (eddy currents are different as well as saturation flux).
    All other things being equal, a highly permeable steel pole pickup is louder than a lowly permeable ALNiCo pole pickup, and both are louder than an air coir pickup, and this is because the permeability of the core will enhance the amount of flux change through the coil, as the string moves nearer and farther from the pole pieces, therefore creating more voltage. The fact that the pole pieces in question also supply a residual flux, which the magnetic guitar string exploits, can be ignored in the analysis. For example, suppose the pickup had no magnet at all, and instead you held a magnet above the pickup, or suppose the string itself was a permanent magnet. The analysis would be the same.

    * actually, for the sake of accuracy, it does change the analysis if the permanent magnet is moved to another location, in that the string would have a different total flux charge, depending on it's distance at any given moment from that permanent magnetic source, but that's a minor point and can be ignored. There hysteresis curve of the steel guitar string would also factor in, but that's getting even more trivial.

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    Last edited by Antigua; 05-20-2018 at 09:36 PM.

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    Last edited by Bagpipe; 05-21-2018 at 04:37 AM.

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    Quote Originally Posted by Bagpipe View Post
    Lets simply omit miller effect here and only talk about additional parallel capacitance. You may or may not be right with your thesis, that you may improve tone a bit with reducing self capacitance. I think we go in accord (all) that we can expect the difference to be very small - probably not to be heard by people like me, or people with a similar ear.
    I haven't read your whole reply yet, and might have to leave for a while, but over on another forum where this stuff is discussed it was generally thought that if the pickup had a lower capacitance, the pickup would remain brighter as the volume was rolled off, since the pickup's capacitance would be isolated with its inductance, which speaks to the value of reduced C, which is on topic for this thread, and you're saying none of that matters vaguely because of Miller effect, so you're not just rebuffing my point, but that made by others, so if you're correct, that would be important, but it would also require thorough reasoning.

    You said " but its influence is usually high enuff to swallow self capacitance of the pickup." I'm not sure what you mean "swallow" the pickup's capacitance. Does it dampen the LC resonance? Change the value of C?

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  30. #65
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    And what happens to miller effect?
    Just for information and maybe lowpass calculation:
    The typical input capacitance of classic (12AX7) tube amps is around 150pF including the Miller effect and internal wiring. This just adds to the cable's (and PU's) capacitance.
    Any active electronics between guitar and amp will isolate from the amp's input.

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    Quote Originally Posted by Bagpipe View Post
    Sorry Antigua, but my opinion varies.

    A highly permeable steel pole is only louder than Alnico's because the Permanent Magnet at the bottom is stronger. And this force is transferred with pretty good efficiency by the steel slugs. The AL Value of AlNiCo isn't very good compared to permanent ferrites. That is why -subjectively- Alnico sounds "weaker" than ceramic+steelcores.
    This is what I mean by you having to admit you're assuming too much. You're stating as fact something which is demonstrably incorrect. AlNiCo poles are much stronger than steel poles with an undermounted magnet. This is evidenced, for example, by "Stratitus" where AlNiCo 5 pulls so hard that it causes the string to lose pitch. This really only ever happed with AlNiCO 5, because it is distinctly strong. I recommend buying one of these so that you can see for yourself https://www.amazon.com/Gaussmeter-Fl...ds=tesla+meter

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    Last edited by Bagpipe; 05-21-2018 at 04:37 AM.

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    Quote Originally Posted by Helmholtz View Post
    Just for information and maybe lowpass calculation:
    The typical input capacitance of classic (12AX7) tube amps is around 150pF including the Miller effect and internal wiring. This just adds to the cable's (and PU's) capacitance.
    Any active electronics between guitar and amp will isolate from the amp's input.
    When you turn the volume control of a 250k pot down to where there is, say, 125k ohm series resistance between the pickup and amplifier, is either the miller effect or the 150pF input C impacting the LC resonance of the guitar pickup?

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    Last edited by Bagpipe; 05-21-2018 at 04:38 AM.

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    Last edited by Bagpipe; 05-21-2018 at 04:37 AM.

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